JEE Advanced PYQ Gravitation – Miscellaneous Mixed Concepts Problems

Q 1 :

Consider a star of mass $m_{2}$ kg revolving in a circular orbit around another star of mass $m_{1}$ kg with $m_{1} ≫ m_{2}$ . The heavier star slowly acquires mass from the lighter star at a constant rate of $γ$ kg/s. In this transfer process, there is no other loss of mass. If the separation between the centers of the stars is $r$ , then its relative rate of change $\frac{1}{r} \frac{d r}{d t}$ (in $s^{- 1}$ ) is given by: [2025]

$- \frac{3 γ}{2 m_{2}}$
$- \frac{2 γ}{m_{2}}$
$- \frac{2 γ}{m_{1}}$
$- \frac{3 γ}{2 m_{1}}$

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(2)

$Here, \frac{m_{2} v^{2}}{r} = \frac{G m_{1} m_{2}}{r^{2}}, or, m_{2} ω^{2} r = \frac{G m_{1} m_{2}}{r^{2}}$

$ω = \sqrt{\frac{G m_{1}}{r^{3}}}$

$L = I ω = m_{2} ω r^{2} = m_{2} \sqrt{\frac{G m_{1}}{r^{3}}} r^{2}$

$L = m_{2} \sqrt{G m_{1} r} = constant$

$l n L = l n m_{2} + l n G + \frac{1}{2} l n m_{1} + \frac{1}{2} l n r$

$0 = \frac{d m_{2}}{m_{2}} + \frac{1}{2} \frac{d m_{1}}{m_{1}} + \frac{1}{2} \frac{d r}{r}$

$\frac{d r}{r d t} = - \frac{2}{m_{2}} \frac{d m_{2}}{d t} - \frac{1}{m_{1}} \frac{d m_{1}}{d t} \approx - \frac{2 γ}{m_{2}}$ $(∵ \frac{d m}{d t} = γ)$

Q 2 :

Consider a spherical gaseous cloud of mass density $ρ (r)$ in free space where $r$ is the radial distance from its center. The gaseous cloud is made of particles of equal mass $m$ moving in circular orbits about the common centre with the same kinetic energy $K$ . The force acting on the particles is their mutual gravitational force. If $ρ (r)$ is constant in time, the particle number density $n (r) = ρ (r) / m$ is
[G is universal gravitational constant] [2019]

$\frac{3 K}{π r^{2} m^{2} G}$
$\frac{K}{2 π r^{2} m^{2} G}$
$\frac{K}{π r^{2} m^{2} G}$
$\frac{K}{6 π r^{2} m^{2} G}$

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(2)

[IMAGE 280]

Gravitational pull of the mass $' M'$ present in the sphere of radius $' r'$ provides the required centripetal force of particle of mass $' m'$ to revolve in a circular path.

$\frac{m v^{2}}{r} = \frac{G M m}{r^{2}} \Rightarrow \frac{1}{2} m v^{2} = \frac{G M m}{2 r} \Rightarrow K = \frac{G M m}{2 r}$

$∴$ $M = \frac{2 K r}{G m}$

$Differentiating the above equation w.r.t. ' r' we get$

$\frac{d M}{d r} = \frac{2 K}{G m}$

$or d M = \frac{2 K}{G m} d r$

$∴$ $4 π r^{2} ρ d r = \frac{2 K}{G m} d r \Rightarrow ρ = \frac{K}{2 π r^{2} m G}$

$∴$ $\frac{ρ}{m} = \frac{K}{2 π r^{2} m^{2} G} or \frac{ρ (r)}{m} = \frac{K}{2 π r^{2} m^{2} G}$

Q 3 :

A thin uniform annular disc (see figure) of mass $M$ has outer radius $4 R$ and inner radius $3 R$ . The work required to take a unit mass from point P on its axis to infinity is [2010]

[IMAGE 281]

$\frac{2 G M}{7 R} (4 \sqrt{2} - 5)$
$- \frac{2 G M}{7 R} (4 \sqrt{2} - 5)$
$\frac{G M}{4 R}$
$\frac{2 G M}{5 R} (\sqrt{2} - 1)$

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(1)

Mass per unit area of the shaded part

$σ = \frac{mass}{area} = \frac{M}{π ({(4 R)}^{2} - {(3 R)}^{2})} = \frac{M}{7 π R^{2}}$

Let us consider a ring of radius $x$ and thickness $d x$ as shown in the figure.

[IMAGE 282]

$Mass of the ring, d M = σ 2 π x d x = \frac{2 π M x d x}{7 π R^{2}}$

Potential at point $P$ due to shaded part

$V_{P} = \int_{3 R}^{4 R} - \frac{G d M}{\sqrt{{(4 R)}^{2} + x^{2}}} = - \frac{G M 2 π}{7 π R^{2}} \int_{3 R}^{4 R} \frac{x d x}{\sqrt{16 R^{2} + x^{2}}}$

Solving, we get

$V_{P} = - \frac{G M 2 π}{7 π R^{2}} {[\sqrt{16 R^{2} + x^{2}}]}_{3 R}^{4 R} = - \frac{2 G M}{7 R} (4 \sqrt{2} - 5)$

Work done in moving a unit mass from $P to \infty = V_{\infty} - V_{P}$

$or W_{P \to \infty} = 0 - (- \frac{2 G M}{7 R} (4 \sqrt{2} - 5)) = \frac{2 G M}{7 R} (4 \sqrt{2} - 5)$

Q 4 :

The distance between two stars of masses $3 M_{s}$ and $6 M_{s}$ is $9 R$ . Here $R$ is the mean distance between the centers of the Earth and the Sun, and $M_{s}$ is the mass of the Sun. The two stars orbit around their common center of mass in circular orbits with period $n T$ , where $T$ is the period of Earth's revolution around the Sun. The value of $n$ is ______. [2021]

(9)

The centre of mass lies at a distance $6 R$ from lighter mass

In circular orbit,

Time Period, $T = 2 π \sqrt{\frac{R^{3}}{G M_{s}}}$

[IMAGE 283]

Binary stars system

$n T = 2 π \sqrt{\frac{{(9 R)}^{3}}{G (3 M_{s} + 6 M_{s})}}$

$or, n \times 2 π \sqrt{\frac{R^{3}}{G M_{s}}} = 9 \times 2 π \sqrt{\frac{R^{3}}{G M_{s}}}$

$[∵ T = 2 π \sqrt{\frac{R^{3}}{G M_{s}}}]$

$∴$ $n = 9$

Q 5 :

A large spherical mass $M$ is fixed at one position and two identical point masses $m$ are kept on a line passing through the centre of $M$ (see figure). The point masses are connected by a rigid massless rod of length $ℓ$ and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to $M$ is at a distance $r = 3 ℓ$ from $M$ , the tension in the rod is zero for $m = k (\frac{M}{288})$ . The value of $k$ is _____ . [2015]

[IMAGE 284]

(7)

For point mass at distance $r = 4 l$

$\frac{G M m}{{(4 l)}^{2}} + \frac{G m^{2}}{l^{2}} = m a$

[IMAGE 285]

For point mass at distance $r = 3 l$

$\frac{G M m}{{(3 l)}^{2}} - \frac{G m^{2}}{l^{2}} = m a$

$∴$ $\frac{G M m}{{(4 l)}^{2}} + \frac{G m m}{l^{2}} = \frac{G M m}{{(3 l)}^{2}} - \frac{G m m}{l^{2}}$

$∴$ $2 m = M [\frac{1}{9} - \frac{1}{16}]$

$\Rightarrow m = \frac{7 M}{288}$

$∴$ $K = 7$

Q 6 :

Two bodies, each of mass $M$ , are kept fixed with a separation $2 L$ . A particle of mass $m$ is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is $G$ . The correct statement(s) is (are) [2013]

The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $4 \sqrt{\frac{G M}{L}}$
The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $2 \sqrt{\frac{G M}{L}}$
The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $\sqrt{\frac{2 G M}{L}}$
The energy of the mass $m$ remains constant

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Select one or more options

(2, 4)

From conservation of mechanical energy,

$\frac{- G M m}{L} - \frac{G M m}{L} + \frac{1}{2} m v^{2} = 0 + 0$

$or, \frac{1}{2} m v^{2} = \frac{2 G M m}{L}$ $∴$ $v = \sqrt{\frac{4 G M}{L}} = 2 \sqrt{\frac{G M}{L}}$

[IMAGE 286]

Total energy of mass $' m'$ is conserved as there is no external force involved.

Topic Question Set

Q 3 :

A thin uniform annular disc (see figure) of mass $M$ has outer radius $4 R$ and inner radius $3 R$ . The work required to take a unit mass from point P on its axis to infinity is [2010]

[IMAGE 281]

Q 6 :

Two bodies, each of mass $M$ , are kept fixed with a separation $2 L$ . A particle of mass $m$ is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is $G$ . The correct statement(s) is (are) [2013]

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Topic Question Set

Q 3 : A thin uniform annular disc (see figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P on its axis to infinity is [2010] [IMAGE 281]

Q 6 : Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G. The correct statement(s) is (are) [2013]

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Q 3 :

A thin uniform annular disc (see figure) of mass $M$ has outer radius $4 R$ and inner radius $3 R$ . The work required to take a unit mass from point P on its axis to infinity is [2010]

[IMAGE 281]

Q 6 :

Two bodies, each of mass $M$ , are kept fixed with a separation $2 L$ . A particle of mass $m$ is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is $G$ . The correct statement(s) is (are) [2013]